Question

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD show that angle A is lesser than angle C and angle B is lesser than angle D​

Answer 1

$\large\underline{\sf{Given \: \: correct \: Question - }}$

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD show that angle A is greater than angle C and angle B is greater than angle D.

$\large\underline{\sf{Solution-}}$

Given that,

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.

Construction :- Join BD

Now, In ∆ ABD

↝ It is given that, AB is the smallest side of quadrilateral ABCD.

So, AD > AB

We know,

Angle opposite to longest side is always greater.

⇛ ∠ABD > ∠ADB ----------- [ 1 ]

Now, In ∆ BCD

↝ It is given that, CD is the longest side of quadrilateral ABCD.

So, CD > CB

⇛ ∠CBD > ∠CDB ------------ [ 2 ]

On Adding equation (1) and (2), we get

⟼ ∠ABD + ∠CBD > ∠ADB + ∠CDB

⇛ ∠ABC > ∠ADC

⇛ ∠B > ∠D

Similarly, ∠A > ∠C

Hence, Proved

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Explore more :-

Properties of a triangle

1. A triangle has three sides, three angles and three vertices.

2. The sum of all internal angles of a triangle is always equal to 180°.

3. The sum of any two sides of a triangle is greater than the third side.

4. The side opposite to the largest angle of a triangle is the largest side.

5. The exterior angle of the triangle is equal to the sum of its interior opposite angles.

6. The angle opposite to longest side is always greater.